Bofors

[Sniper297]

No, I meant YOU were looking at the math wrong.

"And, if it were really true, why would navies even bother with torpedoes and bombs; just use small or medium size guns, and shower the enemy with high volume fire, to do the job"

If it is true that means LESS damage from deck guns and small caliber cannon, not more, so they wouldn't bother with deck guns / Bofors / Oerlikon at all.

"Being pragmatic you can lay waste to the superstructure reducing it to twisted iron but if there are no holes below the WL the ship will not sink."

Depends on what else is in the superstructure and if you can start a fire. I was in bootcamp in 1977, and we were taught that fire is a bigger danger than flooding on a warship. Mainly because of this disaster 10 years before;

http://en.wikipedia.org/wiki/1967_USS_Forrestal_fire

we spent about 3 of the 9 weeks training in firefighting and only 2 days in flooding control.

To me this looked promising;

http://www.subsim.com/radioroom/showthread.php?t=180920

But in my tests it didn't work very well.

We saw this training film;



Bear in mind the ZUNI rocket didn't travel far enough to arm so the warhead didn't explode, but it did make a metal to metal spark which ignited the jet fuel. In addition to the film we also learned firsthand how quickly a fire can spread and get out of control. In WWII they hadn't learned this lesson yet, so only a small percentage of a warship's crew had any training at all in firefighting, and the merchant crews had none.

Again it depends on the cargo, but it should be possible to start a fire with 40mm which would quickly spread and destroy the target, flooding is irrelevant because if it blows up it's gonna sink.

[Crannogman]

So to compare various submarine shells, we would convert their bursting charge into a blast coefficient- per the NavWeaps site, that coefficient is the square root of the bursting charge's mass.

900g 40mm shell charge = 68g, BC = 8.35

5900g 3" shell charge = 340g, BC = 18.44

14970g 4" shell charge = 630g, BC = 25.10

24460g 5" shell charge = 3305g, BC = 57.49

47600g 6" shell charge = 3218g, BC = 56.73

This suggests that the bursting charge from a 5" or 6" gun is equivalent to 7 40mm shells, which seems absurd on the face of it. No doubt the kinetic energy of the entire projectile is also important, and is directly proportional to the shell's mass. So it's not merely a question of whether the square root rule is valid, but what portion of projectile damage is due to kinetic energy and what portion is due to blast effects.

If nobody has any better sources, I might try calling the Naval Surface Warfare Center

[Hambone307]

What I'm gathering is just stick to using my torpedos. All this math is making my head hurt...

[Sniper297]

Tossing all the math out, the ideal would be to set shells.zon and the ammo supply so you can sink one or two small to medium merchant ships per patrol with gunfire alone, leaving torpedoes as your primary weapon.

[TorpX]

Originally Posted by Sniper297

No, I meant YOU were looking at the math wrong.

If you disagree with my ideas, fine, but I KNOW I got the math right.

Again, if small shells are so omnipotent, why not equip cruisers and destroyers with them. They are lighter than 4 or 5 inch guns, and so is the ammunition. It is hard to find examples of significant ships that were sunk (or blown up) with small shells. It is not hard to find examples of ships that were destroyed or sunk with torpedoes and bombs. The reason the small 20mm and 40mm guns were there was for AAA. The reason for the 4 in. guns (at least originally) was to take out small torpedo boats when they tried to attack the big ships.

This suggests that the bursting charge from a 5" or 6" gun is equivalent to 7 40mm shells, which seems absurd on the face of it. No doubt the kinetic energy of the entire projectile is also important, and is directly proportional to the shell's mass. So it's not merely a question of whether the square root rule is valid, but what portion of projectile damage is due to kinetic energy and what portion is due to blast effects.

Certainly, the KE must come into play, but if your 4 in. AP shot produces a 4.2 (?) in. hole, doubling it's KE will not change much. Fragmentation effects are important when it comes to personnel casualties, but are not going to sink ships. Fragments will not have nearly the penetration of an intact shell/shot, though they are deadly to the crews.

[Crannogman]

Doubling the kinetic energy allows the projectile fragments to penetrate farther into/through the target. So maybe you'll get two holes though the hull and maybe on through the boiler to boot. Even the bursting shells had a solid cap that could travel into deeper into the ship.

I guess the major problem with the 40mm is getting hits below the waterline. Still, they might be useful for counterbattery fire.

I use a fire damage mod as well, can't say that it does much

[Sniper297]

Transmitting and receiving on different frequencies or something here.

"I KNOW I got the math right.

Again, if small shells are so omnipotent"

The math they're talking about is REDUCED effectiveness, not INCREASED, rather than the square root making small shells omnipotent, the square root would make them less potent. If the effect was the square of the weight rather than the square root, that would make them more effective.

Again I don't know where the formula came from, but if it's true it would support your opinion that small caliber shells aren't very potent rather than contradicting your opinion. If the math called for the square of the charge then 16 ounces multiplied by 16 would be the equivalent of 256 ounces, more powerful. That's not what it says, it says the square ROOT - the square root of 16 is 4, so a 16 ounce shell is the equivalent of 4 ounces, less powerful. That's assuming the formula is correct, which I have no clue.

That's what's confusing me, you're saying the formula must be wrong because it would make the shells MORE powerful than real life, but the square ROOT would make them LESS powerful?

[Crannogman]

You can't take it based on a single shell - again, the square root is for comparing damage between different bursting charge weights. So, if you double the blasting charge, you only get 1.4x the damage; in order to double the damage, you have to quadruple the weight of the blasting charge. Hence you may perceive diminished returns from increased size of gun and shell.

Even should this square root rule hold true, a larger gun or shell typically delivers more kinetic energy and at longer ranges. Thus, there is still an advantage to carrying larger guns, especially for penetrating armored targets.

[Sniper297]

There is that factor. One of the considerations in the planning of the Pearl Harbor attack was that dive bombing was more accurate, but high level horizontal bombing gave more velocity to the bomb, meaning better armor penetration with the same bomb. Eventually they went with a mix of both high level and dive bombing, plus the torpedo bombers.

[TorpX]

Originally Posted by Sniper297

Transmitting and receiving on different frequencies or something here.

"I KNOW I got the math right.

Again, if small shells are so omnipotent"

The math they're talking about is REDUCED effectiveness, not INCREASED, rather than the square root making small shells omnipotent, the square root would make them less potent. If the effect was the square of the weight rather than the square root, that would make them more effective.

Again I don't know where the formula came from, but if it's true it would support your opinion that small caliber shells aren't very potent rather than contradicting your opinion. If the math called for the square of the charge then 16 ounces multiplied by 16 would be the equivalent of 256 ounces, more powerful. That's not what it says, it says the square ROOT - the square root of 16 is 4, so a 16 ounce shell is the equivalent of 4 ounces, less powerful. That's assuming the formula is correct, which I have no clue.

That's what's confusing me, you're saying the formula must be wrong because it would make the shells MORE powerful than real life, but the square ROOT would make them LESS powerful?

Taking the square root diminishes the difference between types of ordnance. It is not the individual numbers, but the ratios that matter.



If the 'effectiveness' is proportional to the bursting charge itself:

notional small AA shell ______ 0.25 lb. charge _____ 0.25
notional medium shell _______ 1.0 lb. charge ______ 1.0
notional bomb _____________ 100 lb. charge ______ 100.

So, we have a ratio of .25 to 1.0 to 100, (or 1:4:400). That is a bomb is as effective as 100 medium shells or 400 small shells.


If the 'effectiveness' is proportional to the square root of the bursting charge:

notional small AA shell ______ 0.25 lb. charge _____ 0.5
notional medium shell _______ 1.0 lb. charge _____ 1.0
notional bomb _____________ 100 lb. charge _____ 10.0

This gives a ratio of 0.5 to 1.0 to 10.0, (or 1:2:10). Then a bomb is as effective as 10 medium shells or 20 small shells. The ratio of these is, at the extreme, only 20 to 1, where before, it was 400 to 1. If you only need 20 small shells to equal a large bomb, it is more powerful, than if you need 400.

My criticism of this 'square root idea', is that you can get to the point of destroying Yamato-sized ships with glorified heavy machineguns, because small shells can be thrown out in great volumes.

[Sniper297]

Doesn't work that way, instead of decimals convert to ounces.

.25 pound = 4 ounces. Square root of 4 ounces = 2 ounces.
1 pound = 16 ounces. Square root of 16 ounces = 4 ounces.
Square root of 100 pounds is 10 pounds, which is 160 ounces. So the increase over a 1 pound charge would be 40 times the bang, not 10 times the bang. It would be 80 times the bang of a .25 pound (4 ounce) charge.

Again I have no idea if the formula itself is valid, what I'm saying is IF the formula is valid it would make smaller charges less effective rather than more effective per weight.

To me it seems more logical that a 100 pound bomb would have 100 times the bang of a 1 pound bomb, but 40 times the bang wouldn't make the increase meaningless.

[TorpX]

Originally Posted by Sniper297

Doesn't work that way, instead of decimals convert to ounces.

Oh, but it does work that way.

.25 pound = 4 ounces. Square root of 4 ounces = 2 ounces.
1 pound = 16 ounces. Square root of 16 ounces = 4 ounces.
Square root of 100 pounds is 10 pounds, which is 160 ounces. So the increase over a 1 pound charge would be 40 times the bang, not 10 times the bang. It would be 80 times the bang of a .25 pound (4 ounce) charge.

The statement in red isn't valid.

In that line, you took the sq. root first. You didn't do that in the first two lines.

You must first convert the 100 lbs. to ounces, then take the square root (same as you did in the upper 2 lines). Thus, 100 lbs. = 1600 oz. The square root of 1600 is 40. The ratio of 40 oz. to 2 oz. is 20 to 1. This is the same answer you get if you use pounds, as I did.

[Sniper297]

Well, that's why math never made any sense to me, if any formula is valid it should be the same regardless of whether you use grams, ounces, pounds, or tons. Square root doesn't.

2000 pound bomb square root is 44.7, ridiculously low effect which makes me question where the square root theory came from in the first place.

2000 pounds is one ton - square root of 1 is 1, so a 1 ton bomb has the equivalent power of a 1 ton bomb, which is a lot more than 44.7 pounds. A 1000 pound bomb is 1/2 ton, the square root of 0.5 is 0.7, so a half ton bomb has the explosive power of a 3/4 ton bomb? The whole thing is too screwy for words, if you can't convert back and forth and still get the same values the formula just won't work.

[TorpX]

Well, you can use any unit, but you have to use the same unit across the board. That is, you must convert to the selected unit first, then take the root. Otherwise, you end up with a hash.

Basically, this is because when we take the sq. root of a 100 lb. charge, we are creating an arbitrary unit. Instead of pounds, we have sq. root pounds. Just as you can't add linear feet, sq. feet, and cubic feet, you can't mix sq. root lbs., ordinary pounds, and sq. root oz., etc. Defining a function in terms of an arbitrary unit, in itself, doesn't invalidate the idea, but one must be careful to be consistent in the units used.

Where did they get the idea? I suspect they are using a convention from land artillery used in wargames. If you are throwing shells at infantry companies, or troops formations, a bigger bomb doesn't necessarily get you a much better return. The troops are spread out, so you might be better off using a scattering of small mortar shells, instead of one large shell. In this context, the idea makes sense. But, for buildings or ships, I think it is misplaced.

[Sniper297]

That would make sense, the whole reason why rockeyes were invented in the first place. Instead of a single 500 pound mark 82 against a spread out formation you drop a rockeye cluster bomb which spreads 247 1.3 pound bomblets over something like 2500 square yards. Navy used those against small fast boats (kind of a shotgun where you only need one pellet to hit), but they wouldn't be very effective against warships.

"Basically, this is because when we take the sq. root of a 100 lb. charge, we are creating an arbitrary unit. Instead of pounds, we have sq. root pounds. Just as you can't add linear feet, sq. feet, and cubic feet, you can't mix sq. root lbs., ordinary pounds, and sq. root oz., etc. Defining a function in terms of an arbitrary unit, in itself, doesn't invalidate the idea, but one must be careful to be consistent in the units used."

But that's the problem with the whole theory, 2000 pounds is 1 ton is 32,000 ounces, if you divide by 2 or multiply by itself it doesn't matter where you start or where you end, the result will be the same value whenever you convert it. Doesn't work that way with a square root, and if you have to create an arbitrary unit and know what that unit is - and it doesn't work exactly the same whatever unit is used - then the formula is flawed.